QBManager

COORDINATION COMPOUNDS

274157 The shape of ${[{PtCl}_{3} (C_{2} H_{4})]}^{-}$ and the hybridisation of $Pt$ respectively are

1 tetrahedral,

{sp}^{3}

2 trigonal pyramidal,

{sp}^{3}

3 square planar,

{dsp}^{2}

4 square planar,

d^{2} {sp}^{3}

Explanation:

(C) : The complex is ${[{PtCl}_{3} (C_{2} H_{4})]}^{-}$ . The platinum metal connected with the four ligand. The complex formed the square planar geometry because $Pt$ belongs to the $5 d$ series.
$\begin{matrix} {[{PtCl}_{3} (C_{2} H_{4})]}^{-} \\ x + 3 (- 1) + 0 = - 1 \\ x = + 2 \\ {Pt}^{2 +} = [Xe] 5 d^{8} 6 s^{0} 6 p^{0} (Pairing take place) \end{matrix}$
original image
Hence, shape of compound is square planar with ${dsp}^{2}$ hybridization.

UPTU/ UPSEE-2013

COORDINATION COMPOUNDS

274159 Which of the following is an outer orbital complex ?

1

{[Cr {({NH}_{3})}_{6}]}^{3 +}

2

{[Ni {({NH}_{3})}_{6}]}^{2 +}

3

{[Fe (CN)_{5}]}^{3 -}

4

{[Mn (CN)_{6}]}^{4 -}

Explanation:

(B) :
original image

when outer $4 d$ orbitals are involved in hybridisation, it is called an outer orbital complex.

AMU - 2010

COORDINATION COMPOUNDS

274164 In Cu-ammonia complex, the state of hybridization of ${Cu}^{2 +}$ is

1

s p^{3}

2

d^{3} s

3

s p^{2} f

4

d s p^{2}

Explanation:

(D) : Copper form complex with ${NH}_{3}$ in +2 and +3 oxidation state. ${Cu}^{2 +}$ make the square planar complex with ${NH}_{3}$ because ${NH}_{3}$ act as strong field ligand.
${Cu}^{2 +} = [Ar] 3 d^{9} 4 s^{\circ} 4 p^{\circ}$
original image
Hence, the hybridization of ${Cu}^{2 +}$ is ${dsp}^{2}$ .

WB-JEE-2009

COORDINATION COMPOUNDS

274153 Which one of the following complexes is not expected to exhibit isomerism?

1

{[Ni {({NH}_{3})}_{4} {(H_{2} O)}_{2}]}^{2 +}

2

[Pt {({NH}_{3})}_{2} {Cl}_{2}]

3

[Ni {({NH}_{3})}_{2} {Cl}_{2}]

4

{[Ni (en)_{3}]}^{2 +}

Explanation:

(C) : Compounds having tetrahedral geometry does not exhibit isomerism due to presence of symmetry elements. Here, $[Ni {({NH}_{3})}_{2} {Cl}_{2}]$ has tetrahedral geometry.

VITEEE- 2012

COORDINATION COMPOUNDS

274145 According to Werner's theory the geometry of the complex is determined by

1 only from the primary valence in space

2 number and position of the primary valences in space

3 number and position of the secondary valency

4 only from the position of secondary valence in space

MHT CET-02.05.2019

COORDINATION COMPOUNDS

274157 The shape of ${[{PtCl}_{3} (C_{2} H_{4})]}^{-}$ and the hybridisation of $Pt$ respectively are

1 tetrahedral,

{sp}^{3}

2 trigonal pyramidal,

{sp}^{3}

3 square planar,

{dsp}^{2}

4 square planar,

d^{2} {sp}^{3}

Explanation:

(C) : The complex is ${[{PtCl}_{3} (C_{2} H_{4})]}^{-}$ . The platinum metal connected with the four ligand. The complex formed the square planar geometry because $Pt$ belongs to the $5 d$ series.
$\begin{matrix} {[{PtCl}_{3} (C_{2} H_{4})]}^{-} \\ x + 3 (- 1) + 0 = - 1 \\ x = + 2 \\ {Pt}^{2 +} = [Xe] 5 d^{8} 6 s^{0} 6 p^{0} (Pairing take place) \end{matrix}$
original image
Hence, shape of compound is square planar with ${dsp}^{2}$ hybridization.

UPTU/ UPSEE-2013

COORDINATION COMPOUNDS

274159 Which of the following is an outer orbital complex ?

1

{[Cr {({NH}_{3})}_{6}]}^{3 +}

2

{[Ni {({NH}_{3})}_{6}]}^{2 +}

3

{[Fe (CN)_{5}]}^{3 -}

4

{[Mn (CN)_{6}]}^{4 -}

Explanation:

(B) :
original image

when outer $4 d$ orbitals are involved in hybridisation, it is called an outer orbital complex.

AMU - 2010

COORDINATION COMPOUNDS

274164 In Cu-ammonia complex, the state of hybridization of ${Cu}^{2 +}$ is

1

s p^{3}

2

d^{3} s

3

s p^{2} f

4

d s p^{2}

Explanation:

(D) : Copper form complex with ${NH}_{3}$ in +2 and +3 oxidation state. ${Cu}^{2 +}$ make the square planar complex with ${NH}_{3}$ because ${NH}_{3}$ act as strong field ligand.
${Cu}^{2 +} = [Ar] 3 d^{9} 4 s^{\circ} 4 p^{\circ}$
original image
Hence, the hybridization of ${Cu}^{2 +}$ is ${dsp}^{2}$ .

WB-JEE-2009

COORDINATION COMPOUNDS

274153 Which one of the following complexes is not expected to exhibit isomerism?

1

{[Ni {({NH}_{3})}_{4} {(H_{2} O)}_{2}]}^{2 +}

2

[Pt {({NH}_{3})}_{2} {Cl}_{2}]

3

[Ni {({NH}_{3})}_{2} {Cl}_{2}]

4

{[Ni (en)_{3}]}^{2 +}

Explanation:

(C) : Compounds having tetrahedral geometry does not exhibit isomerism due to presence of symmetry elements. Here, $[Ni {({NH}_{3})}_{2} {Cl}_{2}]$ has tetrahedral geometry.

VITEEE- 2012

COORDINATION COMPOUNDS

274145 According to Werner's theory the geometry of the complex is determined by

1 only from the primary valence in space

2 number and position of the primary valences in space

3 number and position of the secondary valency

4 only from the position of secondary valence in space

MHT CET-02.05.2019

COORDINATION COMPOUNDS

274157 The shape of ${[{PtCl}_{3} (C_{2} H_{4})]}^{-}$ and the hybridisation of $Pt$ respectively are

1 tetrahedral,

{sp}^{3}

2 trigonal pyramidal,

{sp}^{3}

3 square planar,

{dsp}^{2}

4 square planar,

d^{2} {sp}^{3}

Explanation:

(C) : The complex is ${[{PtCl}_{3} (C_{2} H_{4})]}^{-}$ . The platinum metal connected with the four ligand. The complex formed the square planar geometry because $Pt$ belongs to the $5 d$ series.
$\begin{matrix} {[{PtCl}_{3} (C_{2} H_{4})]}^{-} \\ x + 3 (- 1) + 0 = - 1 \\ x = + 2 \\ {Pt}^{2 +} = [Xe] 5 d^{8} 6 s^{0} 6 p^{0} (Pairing take place) \end{matrix}$
original image
Hence, shape of compound is square planar with ${dsp}^{2}$ hybridization.

UPTU/ UPSEE-2013

COORDINATION COMPOUNDS

274159 Which of the following is an outer orbital complex ?

1

{[Cr {({NH}_{3})}_{6}]}^{3 +}

2

{[Ni {({NH}_{3})}_{6}]}^{2 +}

3

{[Fe (CN)_{5}]}^{3 -}

4

{[Mn (CN)_{6}]}^{4 -}

Explanation:

(B) :
original image

when outer $4 d$ orbitals are involved in hybridisation, it is called an outer orbital complex.

AMU - 2010

COORDINATION COMPOUNDS

274164 In Cu-ammonia complex, the state of hybridization of ${Cu}^{2 +}$ is

1

s p^{3}

2

d^{3} s

3

s p^{2} f

4

d s p^{2}

Explanation:

(D) : Copper form complex with ${NH}_{3}$ in +2 and +3 oxidation state. ${Cu}^{2 +}$ make the square planar complex with ${NH}_{3}$ because ${NH}_{3}$ act as strong field ligand.
${Cu}^{2 +} = [Ar] 3 d^{9} 4 s^{\circ} 4 p^{\circ}$
original image
Hence, the hybridization of ${Cu}^{2 +}$ is ${dsp}^{2}$ .

WB-JEE-2009

COORDINATION COMPOUNDS

274153 Which one of the following complexes is not expected to exhibit isomerism?

1

{[Ni {({NH}_{3})}_{4} {(H_{2} O)}_{2}]}^{2 +}

2

[Pt {({NH}_{3})}_{2} {Cl}_{2}]

3

[Ni {({NH}_{3})}_{2} {Cl}_{2}]

4

{[Ni (en)_{3}]}^{2 +}

Explanation:

(C) : Compounds having tetrahedral geometry does not exhibit isomerism due to presence of symmetry elements. Here, $[Ni {({NH}_{3})}_{2} {Cl}_{2}]$ has tetrahedral geometry.

VITEEE- 2012

COORDINATION COMPOUNDS

274145 According to Werner's theory the geometry of the complex is determined by

1 only from the primary valence in space

2 number and position of the primary valences in space

3 number and position of the secondary valency

4 only from the position of secondary valence in space

MHT CET-02.05.2019

COORDINATION COMPOUNDS

274157 The shape of ${[{PtCl}_{3} (C_{2} H_{4})]}^{-}$ and the hybridisation of $Pt$ respectively are

1 tetrahedral,

{sp}^{3}

2 trigonal pyramidal,

{sp}^{3}

3 square planar,

{dsp}^{2}

4 square planar,

d^{2} {sp}^{3}

Explanation:

(C) : The complex is ${[{PtCl}_{3} (C_{2} H_{4})]}^{-}$ . The platinum metal connected with the four ligand. The complex formed the square planar geometry because $Pt$ belongs to the $5 d$ series.
$\begin{matrix} {[{PtCl}_{3} (C_{2} H_{4})]}^{-} \\ x + 3 (- 1) + 0 = - 1 \\ x = + 2 \\ {Pt}^{2 +} = [Xe] 5 d^{8} 6 s^{0} 6 p^{0} (Pairing take place) \end{matrix}$
original image
Hence, shape of compound is square planar with ${dsp}^{2}$ hybridization.

UPTU/ UPSEE-2013

COORDINATION COMPOUNDS

274159 Which of the following is an outer orbital complex ?

1

{[Cr {({NH}_{3})}_{6}]}^{3 +}

2

{[Ni {({NH}_{3})}_{6}]}^{2 +}

3

{[Fe (CN)_{5}]}^{3 -}

4

{[Mn (CN)_{6}]}^{4 -}

Explanation:

(B) :
original image

when outer $4 d$ orbitals are involved in hybridisation, it is called an outer orbital complex.

AMU - 2010

COORDINATION COMPOUNDS

274164 In Cu-ammonia complex, the state of hybridization of ${Cu}^{2 +}$ is

1

s p^{3}

2

d^{3} s

3

s p^{2} f

4

d s p^{2}

Explanation:

(D) : Copper form complex with ${NH}_{3}$ in +2 and +3 oxidation state. ${Cu}^{2 +}$ make the square planar complex with ${NH}_{3}$ because ${NH}_{3}$ act as strong field ligand.
${Cu}^{2 +} = [Ar] 3 d^{9} 4 s^{\circ} 4 p^{\circ}$
original image
Hence, the hybridization of ${Cu}^{2 +}$ is ${dsp}^{2}$ .

WB-JEE-2009

COORDINATION COMPOUNDS

274153 Which one of the following complexes is not expected to exhibit isomerism?

1

{[Ni {({NH}_{3})}_{4} {(H_{2} O)}_{2}]}^{2 +}

2

[Pt {({NH}_{3})}_{2} {Cl}_{2}]

3

[Ni {({NH}_{3})}_{2} {Cl}_{2}]

4

{[Ni (en)_{3}]}^{2 +}

Explanation:

(C) : Compounds having tetrahedral geometry does not exhibit isomerism due to presence of symmetry elements. Here, $[Ni {({NH}_{3})}_{2} {Cl}_{2}]$ has tetrahedral geometry.

VITEEE- 2012

COORDINATION COMPOUNDS

274145 According to Werner's theory the geometry of the complex is determined by

1 only from the primary valence in space

2 number and position of the primary valences in space

3 number and position of the secondary valency

4 only from the position of secondary valence in space

MHT CET-02.05.2019

COORDINATION COMPOUNDS

274157 The shape of ${[{PtCl}_{3} (C_{2} H_{4})]}^{-}$ and the hybridisation of $Pt$ respectively are

1 tetrahedral,

{sp}^{3}

2 trigonal pyramidal,

{sp}^{3}

3 square planar,

{dsp}^{2}

4 square planar,

d^{2} {sp}^{3}

Explanation:

(C) : The complex is ${[{PtCl}_{3} (C_{2} H_{4})]}^{-}$ . The platinum metal connected with the four ligand. The complex formed the square planar geometry because $Pt$ belongs to the $5 d$ series.
$\begin{matrix} {[{PtCl}_{3} (C_{2} H_{4})]}^{-} \\ x + 3 (- 1) + 0 = - 1 \\ x = + 2 \\ {Pt}^{2 +} = [Xe] 5 d^{8} 6 s^{0} 6 p^{0} (Pairing take place) \end{matrix}$
original image
Hence, shape of compound is square planar with ${dsp}^{2}$ hybridization.

UPTU/ UPSEE-2013

COORDINATION COMPOUNDS

274159 Which of the following is an outer orbital complex ?

1

{[Cr {({NH}_{3})}_{6}]}^{3 +}

2

{[Ni {({NH}_{3})}_{6}]}^{2 +}

3

{[Fe (CN)_{5}]}^{3 -}

4

{[Mn (CN)_{6}]}^{4 -}

Explanation:

(B) :
original image

when outer $4 d$ orbitals are involved in hybridisation, it is called an outer orbital complex.

AMU - 2010

COORDINATION COMPOUNDS

274164 In Cu-ammonia complex, the state of hybridization of ${Cu}^{2 +}$ is

1

s p^{3}

2

d^{3} s

3

s p^{2} f

4

d s p^{2}

Explanation:

(D) : Copper form complex with ${NH}_{3}$ in +2 and +3 oxidation state. ${Cu}^{2 +}$ make the square planar complex with ${NH}_{3}$ because ${NH}_{3}$ act as strong field ligand.
${Cu}^{2 +} = [Ar] 3 d^{9} 4 s^{\circ} 4 p^{\circ}$
original image
Hence, the hybridization of ${Cu}^{2 +}$ is ${dsp}^{2}$ .

WB-JEE-2009

COORDINATION COMPOUNDS

274153 Which one of the following complexes is not expected to exhibit isomerism?

1

{[Ni {({NH}_{3})}_{4} {(H_{2} O)}_{2}]}^{2 +}

2

[Pt {({NH}_{3})}_{2} {Cl}_{2}]

3

[Ni {({NH}_{3})}_{2} {Cl}_{2}]

4

{[Ni (en)_{3}]}^{2 +}

Explanation:

(C) : Compounds having tetrahedral geometry does not exhibit isomerism due to presence of symmetry elements. Here, $[Ni {({NH}_{3})}_{2} {Cl}_{2}]$ has tetrahedral geometry.

VITEEE- 2012

COORDINATION COMPOUNDS

274145 According to Werner's theory the geometry of the complex is determined by

1 only from the primary valence in space

2 number and position of the primary valences in space

3 number and position of the secondary valency

4 only from the position of secondary valence in space

MHT CET-02.05.2019

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